A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a m

Q: Why does the maximum energy correspond to the transition to the ground state?

The energy of a photon emitted during a transition is the difference between the initial and final energy levels. Since the ground state ($n=1$) is the lowest possible energy level (most negative), dropping all the way down to it releases the maximum possible energy.

Q: Why did we divide the two equations?

We had a system of two non-linear equations with two variables, $E_1$ and $n$. By dividing them, the $E_1$ terms perfectly cancel out, leaving us with a single equation entirely in terms of $n$, which is much easier to solve.

Q: How do we know the minimum energy transition is from $2n$ to $2n-1$?

Energy levels in a hydrogen-like atom get closer together as the quantum number increases. Therefore, the smallest energy gap from any state $k$ is always to the immediately adjacent lower state, $k-1$. In our case, from state $2n$, the smallest jump is to $2n-1$.

Q: Why is the ground state energy $E_1$ negative?

The negative sign indicates that the electron is bound to the nucleus. An energy of $0\text{ eV}$ corresponds to an electron that is completely free (at an infinite distance). To free the electron from the ground state, you must supply positive energy equal to the magnitude of $E_1$.

Q: Can $n$ be a fraction or a decimal?

No, $n$ represents the principal quantum number, which must always be a positive integer ($1, 2, 3, \dots$). If your algebraic solution for $n$ yields a fraction or a negative number, it means there's an error in the setup or calculation.

Question

A hydrogen like atom of atomic number $Z$ is in an excited state of quantum number $2n$. It can emit a maximum energy photon of $204	ext{ eV}$. If it makes a transition to quantum state $n$, a photon of energy $40.8	ext{ eV}$ is emitted. Find $n$, $Z$ and the ground state energy (in $	ext{eV}$) of this atom. Also, calculate the minimum energy (in $	ext{eV}$) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is $-13.6	ext{ eV}$.

Sigma Percentile · Accepted Answer

The journey into the quantum realm of a hydrogen-like atom is always filled with elegant algebraic symmetries and profound physical insights. In this problem, we are tasked with acting as quantum detectives, piecing together the identity and properties of an unknown atom based solely on the light it emits. Let's dive into the thought process that unravels this mystery.

### Decoding the Energy Levels

We begin with the foundational principle of Bohr's model for hydrogen-like atoms. The energy of an electron in the $k^{	ext{th}}$ orbit is inversely proportional to the square of the principal quantum number, $k$. We can express this elegantly as:
$$E_k = \frac{E_1}{k^2}$$
where $E_1$ is the ground state energy of the atom. This simple relationship is the master key that will unlock the entire problem.

### Setting Up the Transitions

The problem provides us with two crucial pieces of evidence regarding the photons emitted during de-excitation from the state $2n$.

First, we are told that the **maximum energy** photon emitted is $204	ext{ eV}$. In any atomic system, the maximum energy is released when the electron makes the largest possible jump—from its current state all the way down to the ground state ($n=1$).
Mathematically, this transition is from $2n ightarrow 1$:
$$E_{2n} - E_1 = 204$$
Substituting our energy formula, we get:
$$\frac{E_1}{(2n)^2} - E_1 = 204 \implies E_1 \left( \frac{1}{4n^2} - 1 ight) = 204$$

Second, we are given the energy of an intermediate transition. When the electron drops from state $2n$ to state $n$, it emits a photon of $40.8	ext{ eV}$.
This transition is from $2n ightarrow n$:
$$E_{2n} - E_n = 40.8$$
Again, substituting the energy formula:
$$\frac{E_1}{(2n)^2} - \frac{E_1}{n^2} = 40.8$$
Taking a common denominator, this simplifies beautifully:
$$E_1 \left( \frac{1 - 4}{4n^2} ight) = 40.8 \implies E_1 \left( \frac{-3}{4n^2} ight) = 40.8$$

### The Algebraic Elegance

We now have a system of two equations with two unknowns, $E_1$ and $n$. The most elegant way to solve this is to divide the first equation by the second. Notice how the $E_1$ and the $4n^2$ terms perfectly cancel out, leaving us with a simple equation in terms of $n$:
$$\frac{E_1 \left( \frac{1 - 4n^2}{4n^2} ight)}{E_1 \left( \frac{-3}{4n^2} ight)} = \frac{204}{40.8}$$
$$\frac{1 - 4n^2}{-3} = 5$$
Multiplying both sides by $-3$ gives:
$$1 - 4n^2 = -15 \implies 4n^2 = 16 \implies n^2 = 4$$
Since the principal quantum number must be a positive integer, we find that **$n = 2$**.

### Unveiling the Atom

With $n$ in hand, we can substitute it back into our second equation to find the ground state energy, $E_1$:
$$E_1 \left( \frac{-3}{4(2)^2} ight) = 40.8 \implies E_1 \left( \frac{-3}{16} ight) = 40.8$$
Solving for $E_1$, we get:
$$E_1 = 40.8 	imes \left( \frac{-16}{3} ight) = \mathbf{-217.6	ext{ eV}}$$

Now, we can identify the atom. We know that the ground state energy of a hydrogen-like atom scales with the square of its atomic number, $Z$:
$$E_1 = -13.6 Z^2$$
Setting this equal to our calculated $E_1$:
$$-217.6 = -13.6 Z^2 \implies Z^2 = \frac{-217.6}{-13.6} = 16$$
Thus, **$Z = 4$**. The atom is a Beryllium ion ($	ext{Be}^{3+}$)!

### The Final Leap: Minimum Energy

Finally, we must calculate the minimum energy emitted during de-excitation from the state $2n$. Since $n=2$, the atom is initially in the $4^{	ext{th}}$ state. The minimum energy corresponds to the smallest possible jump, which is to the immediately adjacent lower state, $4 ightarrow 3$.
$$E_{	ext{min}} = E_4 - E_3 = \frac{E_1}{4^2} - \frac{E_1}{3^2}$$
$$E_{	ext{min}} = E_1 \left( \frac{1}{16} - \frac{1}{9} ight) = E_1 \left( \frac{9 - 16}{144} ight) = E_1 \left( \frac{-7}{144} ight)$$
Substituting our value for $E_1$:
$$E_{	ext{min}} = (-217.6) \left( \frac{-7}{144} ight) = \mathbf{10.58	ext{ eV}}$$
And with that, the quantum mystery is fully resolved!

Visualized Solution

$E_{k}$

$E_{2 n} \to E_{1}$

$Eq. 1$

$E_{2 n} \to E_{n}$

$Eq. 2$

$Ratio$

$n$

$E_{1}$

$Z$

$E_{min}$

$E_{min} Eq.$

$E_{min} Ans.$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Decoding the Energy Levels

Setting Up the Transitions

The Algebraic Elegance

Unveiling the Atom

The Final Leap

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Visualized Solution

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